When 400 mL of 0.2 M H_2SO_4 is mixed with | vedclass

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(A) $1$. Calculate the moles of reactants: $n(H_2SO_4) = 0.4 \ L 	imes 0.2 \ M = 0.08 \ mol$. $n(NaOH) = 0.6 \ L 	imes 0.1 \ M = 0.06 \ mol$. $2$. T

Vedclass · Accepted Answer

$0.82$

Vedclass · Answer

(A) $1$. Calculate the moles of reactants: $n(H_2SO_4) = 0.4 \ L 	imes 0.2 \ M = 0.08 \ mol$. $n(NaOH) = 0.6 \ L 	imes 0.1 \ M = 0.06 \ mol$. $2$. The reaction is $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$. Since $NaOH$ is the limiting reagent, the heat evolved corresponds to the neutralization of $0.06 \ mol$ of $NaOH$. $3$. The total volume of the solution is $400 \ mL + 600 \ mL = 1000 \ mL = 1 \ L$. Assuming the density of the solution is $1 \ g/mL$, the mass of the solution is $1000 \ g$. $4$. Using the formula $q = m 	imes c 	imes \Delta T$: $3430 \ J = 1000 \ g 	imes 4.18 \ J \ K^{-1} \ g^{-1} 	imes \Delta T$. $\Delta T = \frac{3430}{4180} \approx 0.82 \ K$.

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